x'(r)=r*3r5e32((rcr*)2-(rcr)2)[1-(r*r)6e3((rcr*)2-(rcr)2)]-12,

where the constant is chosen to be \(\frac{\sqrt{r^6 e^{3\left(\frac{r_c}{r}\right)^2}-{r^*}^6 e^{3(\frac{r_c}{r^*})^2}}}{r^2}\) . Plugging eq.(REF ) back into eq.(REF ), we find

A=2r*2L2r*1 u-3 e32(rcr*)2u2[1-u6e3(rcr*)2(1-u2)]-12du,

where \(u=\frac{r^*}{r}\)  and \(\epsilon \)  is an ultraviolet cut off. By integrating the differential equation (REF ), the relation between \(l\)  and \(r^*\)  is given by

l=2r*01 u3 e32(rcr*)2(1-u2)[1-e3(rcr*)2(1-u2)u6]-12 du.

Unfortunately eqs. (REF ) and (REF ) can not be analytically solved. Therefore, we need to use the following binomial expansion [1]}, [2]}

(1+x)-r=n=0(-1)nr+n-1nxn ,        |x|<1,

where \(x\)  and \(r\)  are real numbers and \(r>0\) . By identifying \(x=-e^{3\left(\frac{r_c}{r^*}\right)^2(1-u^2)}u^6\)  and using the facts that \(0<u<1\)  and \(r_c<r^*\) , one can easily see that \(\vert x\vert <1\)  and hence the sum is well-defined. Using eq.(REF ), we can write eqs. (REF ) and (REF ) as follows

l=2r*n=0(n+12)(n+1)01 u6n+3 e3(n+12)(rcr*)2(1-u2) du,


A=2r*2L2n=0(n+12)(n+1)r*1 u6n-3e(3n+(32-3n)u2)(rcr*)2 du.

In order to find the \(\mathcal {A}\)  as a function of \(l\) , we should solve eq.(REF ) for \(r^*\)  and then substitute back this in eq.(REF ). Since we can not analytically solve the eq.(REF ) to find \(r^*\)  as a function of \(l\) , we need to focus on the high energy limit (\(l\Lambda _c\ll 1\) ). Note that, in the low energy limit, i.e. \(l\Lambda _c\gg 1\) , or equivalently \(r^*\rightarrow r_c\)  we do not reach the analytical results and then we have to neglect studying this limit.
